Question 145168
Find the eccentricity of the ellipse given by 
{{{16x^2+25y^2=100}}}


I have tried:
{{{(16x^2)/100 + (25y^2)/100 = 1}}} 


that reduces to:
{{{4x^2/25 + 1y^2/4 =1}}}

What do I do next? Please explain if you can! Thanks so very much!
<pre><font size = 4 color = "indigo"><b>
Your error is in thinking that you are necessarily reducing the 
fractions to lowest terms.  Instead think of it as making the 
numerators 1.  Sometimes it amounts to reducing the fraction
but not always.

Do it this way instead. Go back to:

{{{16x^2/100 + 25y^2/100 = 1}}}

Get a {{{1}}} coefficient where the {{{16}}} is by 
multiplying top and bottom of the first fraction by
{{{(1/16)}}}, and get a {{{1}}} coefficient where 
the {{{25}}} is on the second fraction by multiplying 
top and bottom by {{{(1/25)}}}, and we have this:

{{{( 16x^2(1/16) )/(100(1/16)) + (25y^2(1/25))/(100(1/25)) = 1}}}

Now cancel and that leaves just 1 understood on top:

{{{( cross(16)x^2(1/cross(16)) )/(100(1/16)) + (cross(25)y^2(1/cross(25)))/(100(1/25)) = 1}}}

So we have

{{{x^2 /(100/16) + y^2/(100/25) = 1}}}

And we only need to reduce the fractions on the bottom

{{{x^2 /(25/4) + y^2/4 = 1}}}

The larger denominator is {{{a^2}}}, and since {{{25/4>4}}},

we compare the above to:

{{{x^2/a^2 + y^2/b^2=1}}}

So 

{{{a^2=25/4}}}            {{{b^2=4}}}
{{{a=sqrt(25/4)}}}        {{{b=2}}} 
{{{a=5/2}}}

Eccentricity of an ellipse = {{{c/a}}} where {{{c^2=a^2-b^2}}}

{{{c^2=a^2-b^2}}}
{{{c^2=(5/2)^2-2^2}}}
{{{c^2=25/4-4}}}
{{{c^2=25/4-4/1}}}
{{{c^2=25/4-(4*4)/(1*4)}}}
{{{c^2=25/4-16/4}}}
{{{c^2=(25-16)/4}}}
{{{c^2=9/4}}}
{{{c=sqrt(9/4)}}}
{{{c=3/2}}}

Eccentricity of this elipse = {{{c/a = (3/2)/(5/2)= (3/2)(2/5)= (3/cross(2))(cross(2)/5)= 3/5}}}
  
Edwin</pre>